等差数列{an},{bn}的前n项和分虽为Sn和Tn,若Sn/Tn=2n/3n+1,则a8/b8= ,an/bn=

等差数列,所以前n项和为an^2+bn+c的函数.Sn/Tn=2n/3n+1=2n^2/(3n^2+n),可设sn=2n^2,Tn=3n^2+n则an=sn-s(n-1)=2n^2-2(n-1)^2bn=Tn-T(n-1)=3n^2+n-3(n-1)^2-n+1=3n^2+1-3(n-1)^2a8/b8=[ 2n^2-2(n-1)^2]/[3n^2+1-3(n-1)...